Diffeomorphisms of the torus $\mathbb T^n$ While reading the first chapter of Orbifolds and Stringy Topology by Adem, Leida and Ruan, I stumbled upon the following assertion, in page 5

The group
$\text{GL}_n(\mathbb Z)$ acts by matrix multiplication on $\mathbb R^n$, taking the lattice $\mathbb Z^n$ to itself. This
then induces an action on $\mathbb T^n=(\mathbb R/\mathbb Z)^n$. In fact, one can easily show that the
map induced by looking at the action in homology, $\Phi:\text{Aut}(\mathbb T^n)\rightarrow\text{GL}_n(\mathbb Z)$,
is a split surjection.

What? I know that a linear isomorphism $A:\mathbb R^n\rightarrow\mathbb R^n$ induces a diffeomorphism of $\mathbb T^n$ if, and only if the matrix associated to $A$ has coefficients in $\mathbb Z$ and has determinant equal to $\pm1$, that is, if it belongs to $\text{GL}_n(\mathbb Z)$. There are many posts about that on this site, but I am puzzled about that action in homology thing, and that it induces an split surjection from the group  $\text{Aut}(\mathbb T^n)$ onto $\text{GL}_n(\mathbb Z)$. I can imagine diffeomorphisms of the torus that don't come from any linear isomorphism on $\mathbb R^n$, but why should these have any structure whatsoever?
Thanks in advance for your answer.
 A: What it means is the following.
For any continuous map $\phi:\Bbb T^n\rightarrow \Bbb T^n$, there is an induced homomorphism of groups: $\phi_*: H_1(\Bbb T^n) \rightarrow H_1(\Bbb T^n)$.
We know that $H_1(\Bbb T^n)$ is isomorphic to $\Bbb Z^n$. Thus we can view $\phi_*$ as an endomorphism of $\Bbb Z^n$.
Now suppose that $\phi \in \operatorname{Aut}(\Bbb T^n)$. This means that it is not only a continuous map, but also has a continuous inverse. Therefore we see that $\phi_*$ is an automorphism of the group $\Bbb Z^n$, i.e. $\phi_*\in \operatorname{GL}_n(\Bbb Z)$.
By functoriality, the map $\phi \mapsto \phi_*$ is a group homomorphism from $\operatorname{Aut}(\Bbb T^n)$ to $\operatorname{GL}_n(\Bbb Z)$. This is the $\Phi$ map in your question.
They then state that this map is a split surjection. This is equivalent to sayting that there exists a splitting homomorphism $\Psi: \operatorname{GL}_n(\Bbb Z)\rightarrow \operatorname{Aut}(\Bbb T^n)$ such that $\Phi \circ \Psi$ is the identity map on $\operatorname{GL}_n(\Bbb Z)$.
This splitting map $\Psi$ is just constructed as you have described: for any $g\in \operatorname{GL}_n(\Bbb Z)$, the map $x \mapsto gx$ defined on $\Bbb R^n$ induces a map from $\Bbb T^n$ to itself, which we define as $\Psi(g)$.
